If A and B are two events such that P A = 1 3 , P B = 1 5 and P A ∪ B = 1 2 , then P A B ' + P B A '…

Given,

$P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{5}$ and $P\left(A\cup B\right)=\frac{1}{2}$

We know that,

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$\Rightarrow P\left(A\cap B\right)=\frac{1}{3}+\frac{1}{5}-\frac{1}{2}=\frac{1}{30}$

Now $P\left(A\cap B^{\text{'}}\right)=\text{Only A}=\frac{1}{3}-\frac{1}{30}=\frac{9}{30}$

And $P\left(B\cap A^{\text{'}}\right)=\text{Only B}=\frac{1}{5}-\frac{1}{30}=\frac{5}{30}$

Now, $P\left(A\mid B^{\text{'}}\right)+P\left(B\mid A^{\text{'}}\right)=\frac{P\left(A\cap B^{\text{'}}\right)}{P\left(B^{\text{'}}\right)}+\frac{P\left(B\cap A^{\text{'}}\right)}{P\left(A^{\text{'}}\right)}$

$=\frac{\frac{9}{30}}{\frac{4}{5}}+\frac{\frac{5}{30}}{\frac{2}{3}}=\frac{5}{8}$